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Q.

Consider the line L given by the equation $\frac{x - 3}{2} = \frac{y - 1}{1} = \frac{z - 2}{1}$ . Let Q be the mirror image of the point (2, 3, −1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?

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a

(-1, 1, 2)

b

(1, 2, 2)

c

(1, 1, 2)

d

(1, 1, 1)

answer is D.

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Detailed Solution

Given, line

$\Rightarrow \frac{x - 3}{2} = \frac{y - 1}{1} = \frac{z - 2}{1}$ …(i)

Let plane P be a passing through Q and is perpendicular to line (i), then equation of plane be

$\begin{array}{l} 2 (x - 2) + 1 (y - 3) + 1 (z + 1) = 0 \\ 2 x + y + z - 6 = 0 \end{array}$

Now, from options given only point (1, 2, 2) satisfies it.

$∴ (1, 2, 2)$ point is on plane P.

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Consider the line L given by the equation x−32=y−11=z−21. Let Q be the mirror image of the point (2, 3, −1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?